The complete combustion of the alkane required 11.2 L of O2, while 6.72 L of CO2 was released

The complete combustion of the alkane required 11.2 L of O2, while 6.72 L of CO2 was released. Determine the alkane formula.

Given:
CnH2n + 2 – alkane
V (O2) = 11.2 l
V (CO2) = 6.72 l
Find: Formula alkane-?
Solution:
Let’s compose the reaction equation:
CxHy + O2 = CO2 + H2O
v = V / Vm
v (O2) = 11.2 / 22.4 = 0.5 mol.
2) v (CO2) = 6.72 / 22.4 = 0.3 mol.
CO2 deficient
v (CO2) = v (H2O) = 0.3 mol
3) m (CO2) = 0.3 * (12 + 32) = 13.2 g.
m (H2O) = 0.3 * 18 = 5.4 g.
4) 18 g H2O – 2 g H2
5.4 g H2O – y g H2
y = 5.4 * 2/18 = 0.6
5) 44 g CO2 – 12 g C
13.2 g H2O – x g C
x = 13.2 * 12/44 = 3.6
6) C: H = 3.6 / 12: 0.6
C: H = 0.3: 0.6
C: H = 1: 2
Alkane – CH4 – methane

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